Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²): Question bank

1.3.5 · D5 · Physics › Work, Energy & Power › Potential energy — definition, gravitational (mgh and −GMm -

True or false — justify karo

Gravitational PE surface ke paas negative ho sakti hai.

True —

m g h

kisi bhi point ke liye negative hoga jo tumhare chosen

h = 0

line se neeche ho, jaise kisi kuan ki tali. Sign bilkul iss baat par depend karta hai ki tumne reference kahan rakha. Dekho Work done by a force.

Kya formula

- GM m / r

kabhi positive ho sakta hai.

False —

G

M

m

r

sab positive hain aur ek explicit minus sign bhi hai, isliye

- GM m / r < 0

hamesha. Ye sirf zero approach karta hai (neeche se) jab

r \to \infty

Ek akela isolated object potential energy rakhta hai.

False — PE do interacting bodies ke beech ke configuration mein hoti hai (book+Earth, ball+spring). "Book ki PE" sirf book–Earth system ki PE ka shorthand hai.

Spring ko utni hi matra compress karna, stretch karne se kam energy store karta hai.

False —

U = \frac{1}{2} k x^{2}

x^{2}

par depend karta hai, isliye

U (+ x) = U (- x)

. 5 cm push in karo ya 5 cm pull out karo: stored energy bilkul same hogi.

Friction ka ek potential energy function hota hai bilkul gravity ki tarah.

False — friction ka kaam path-dependent hota hai (ek loop karke wapas aane par har cm mein energy lost hoti hai), isliye koi single-valued

U

exist nahi karta. Sirf conservative forces ki PE hoti hai.

Height double karne par near-surface gravitational PE double ho jaati hai.

True —

U = m g h

h

mein linear hai, isliye

h \to 2 h

deta hai

U \to 2 U

. (Spring se compare karo, jo quadratic hai.)

Spring ki stretch double karne par stored energy double ho jaati hai.

False —

U = \frac{1}{2} k x^{2}

quadratic hai, isliye

x \to 2 x

deta hai

U \to 4 U

. Chaar guna energy, do guna nahi.

Jab

r \to \infty

, gravitational PE

- GM m / r

zero ho jaati hai.

True — yahi exactly chosen reference hai:

U = 0

infinite separation par. Mass "well ki rim" par baitha hota hai.

Sabse gehri (most negative) gravitational PE sabse chhote

r

par hoti hai.

True — jaise

r

shrink hota hai,

- GM m / r

ek bada negative number ban jaata hai, isliye mass well mein aur gehra baithta hai aur bahar nikalne ke liye zyada energy chahiye.

Agar PE badhti hai, to conservative force ne positive work kiya hoga.

False — bilkul ulta.

Δ U = - W_{cons}

, isliye

U

badhna matlab

W_{cons} < 0

; force ne motion ko resist kiya jabki tumne iske against kaam kiya.

Error dhundo

"Spring PE

k x^{2}

hai kyunki force

k x

hai."

Force

k x

ko integrate karna padega, sirf multiply nahi:

\int_{0}^{x} k x^{'} d x^{'} = \frac{1}{2} k x^{2}

. Missing

\frac{1}{2}

ek linearly-rising force ka average factor hai. Dekho Hooke's Law.

m g h

aur

- GM m / r

agree nahi karte, isliye ek galat hai."

Dono mein se koi galat nahi hai —

m g h

- GM m / r

ka flat-Earth limit hai jab

h ≪ R

, jisse

Δ U \approx GM m h / R^{2} = m g h

milta hai,

g = GM / R^{2}

use karke. Dono differences par agree karte hain, aur yahi physically matter karta hai.

"Earth se escape karne ki energy

\frac{1}{2} GM m / R

hai kyunki ye average PE hai."

Ye poori well depth hai,

U_{\infty} - U_{R} = 0 - (- GM m / R) = GM m / R

U = 0

rim tak pahunchne ke liye puri depth supply karni padegi; dekho Escape velocity.

F = - d U / d x

: kyunki

U = \frac{1}{2} k x^{2}

hamesha positive hai, force bhi hamesha positive hai."

U

ka sign irrelevant hai; force slope se aata hai.

F = - \frac{d}{d x} (\frac{1}{2} k x^{2}) = - k x

, jo

x

ke saath sign flip karta hai — dono taraf restoring hota hai. Dekho Force from potential — F = -dU/dx.

"Kyunki gravity neeche kheenchti hai, book uthana uss par negative work karta hai, isliye

Δ U

negative hai."

Gravitational force negative work karta hai (

W_{grav} = - m g h

), lekin

Δ U = - W_{grav} = + m g h > 0

Δ U = - W_{cons}

mein minus sign exactly yahi karta hai ki storage positive nikle.

"Shelf ke upar PE 29.4 J hai, isliye book mein 29.4 J hai."

Ye number tumhare chosen floor reference se difference hai, koi absolute content nahi. Reference ko shelf par shift karo aur same book 0 J padhe gi.

"Natural length par ek spring mein zero force aur zero PE hoti hai, isliye ye dono baatein same statement hain."

Ye sirf isliye coincide karte hain kyunki

\frac{1}{2} k x^{2}

ka slope aur height dono

x = 0

par vanish karte hain. Generally zero force (

d U / d x = 0

)

U

ka ek flat point mark karta hai, jo kisi bhi nonzero PE value par ho sakta hai.

Why questions

Δ U = - W_{cons}

mein minus sign kyun hai?

Force dwara positive work stored energy kharchta hai isliye

U

girta hai; force ke against kiya gaya kaam energy bank karta hai isliye

U

badhta hai. Minus "force dwara kaam" ko "stored energy" mein convert karta hai.

Sirf conservative forces ko hi potential energy kyun milti hai?

Unka kaam do points ke beech path-independent hota hai, isliye

U

position ka single-valued function ho sakta hai. Ek path-dependent force ek point ko kai alag

U

values assign karega — jo ek contradiction hai. Dekho Conservative vs non-conservative forces.

Bound gravitational PE negative kyun hoti hai, positive kyun nahi?

Reference

U = 0

infinity par baitha hai; jaise mass andar girta hai gravity positive work karta hai,

U

ko zero se neeche "well" mein dhakelta hai. Negative ka matlab sirf "infinitely-far reference se neeche" hai.

Spring PE

x

mein quadratic kyun hai lekin gravitational

m g h

sirf

h

mein linear kyun hai?

Spring force

k x

displacement ke saath badhta hai, isliye ek rising line ko integrate karne par quadratic milta hai. Near-surface gravity

m g

constant hai, isliye ek flat force ko integrate karne par straight line milti hai.

Force lower PE ki taraf kyun point karta hai?

F_{x} = - d U / d x

matlab force energy landscape ki sabse steep slope ko neeche dhakelta hai — jaise ek ball valley mein roll karta hai. Giravahat jitni steep, push utna zyada.

Ground ke paas varying

- GM m / r^{2}

ko ignore karke constant

m g

kyun use kar sakte hain?

h ≪ R

heights par distance negligible fraction badlta hai, isliye

g = GM / r^{2}

barely change hota hai. Uniform-field approximation sirf

h / R

ke level par galti karta hai. Dekho Gravitation — Newton's law.

PE system ki property kyun hai, kisi ek object ki kyun nahi?

PE do interacting bodies ki relative configuration measure karta hai; ek ko move karo aur stored work badal jaata hai. Sirf ek body ke saath koi conservative force ke beech kuch nahi hota. Linked to Conservation of mechanical energy.

Infinity tak escape karne ke liye exactly

GM m / R

kyun chahiye, zyada kyun nahi?

Kyunki

U = 0

infinity par define hai, isliye poori chadhti well depth ke barabar hai

0 - (- GM m / R) = GM m / R

. Koi extra energy leftover kinetic energy ke roop mein pahunche gi, climbing cost ke roop mein nahi.

Edge cases

r = 0

par gravitational PE

- GM m / r

kya hogi?

- \infty

par diverge karta hai — ek mathematical signal ki true point mass idealized hai; real bodies ki finite radius hoti hai, isliye tum kabhi unke bahar

r = 0

tak nahi pahunchte.

Natural length par spring ki PE kya hogi?

Hamare chosen reference

x = 0

se zero, aur ye

U = \frac{1}{2} k x^{2}

ka minimum hai, isliye force

- k x

bhi wahan vanish hota hai — ek stable equilibrium.

Agar tum PE reference floor ki jagah shelf par rakho, to kya book ki motion badlegi?

Nahi — physics mein sirf

Δ U

enter karta hai, aur differences reference shift karne se unchanged rehte hain. Predicted speeds aur forces bilkul same honge.

Agar ek point reference se neeche ho (

h < 0

) to

m g h

ka kya hoga?

U = m g h

negative ho jaata hai, theek se batata hai ki object reference ke relative ek valley mein baitha hai. Formula

h

ke dono signs handle karta hai bina kisi badlaav ke.

Throw kiye gaye ball ki flight ke exact top par, kya uski gravitational PE maximum hai ya force zero?

PE maximum hai (sabse uncha point), lekin force zero nahi hai — gravity

m g

tab bhi neeche kheenchti hai; yahi motion ko reverse karta hai. Max PE ≠ zero force yahan kyunki

g

constant hai, slope-dependent nahi.

Do masses infinitely door par hone par, system ki gravitational PE aur force kya hogi?

Dono zero approach karte hain:

U = - GM m / r \to 0

aur

F = - GM m / r^{2} \to 0

. Infinitely alag bodies ek dusre ko barely feel karti hain — natural "kuch stored nahi, kuch kheench nahi raha" state.

Kya bound orbit mein total mechanical energy negative ho sakti hai?

Haan — agar (negative) gravitational PE kinetic energy se zyada ho,

E = K + U < 0

, jo exactly ek bound, non-escaping orbit ki pehchaan hai. Dekho Conservation of mechanical energy.

Recall Jaane se pehle ek-line self-test

Inhe cover karo aur reason ke saath jawab do. Bound gravitational PE ka sign aur kyun? ::: Negative, kyunki reference infinity par hai aur andar girne par positive work hota hai, $U$ ko zero se neeche le jaata hai. Spring PE mein $\frac{1}{2}$ kyun? ::: Ye linearly-rising force $k x$ ko integrate karne ka average factor hai. PE kiska property hai? ::: Interacting system ka, kisi ek object ka kabhi nahi.